Problem: Simplify; express your answer in exponential form. Assume $n\neq 0, r\neq 0$. $\dfrac{{(n^{-1})^{-4}}}{{(n^{-5}r^{3})^{-5}}}$
Answer: To start, try working on the numerator and the denominator independently. In the numerator, we have ${n^{-1}}$ to the exponent ${-4}$ . Now ${-1 \times -4 = 4}$ , so ${(n^{-1})^{-4} = n^{4}}$ In the denominator, we can use the distributive property of exponents. ${(n^{-5}r^{3})^{-5} = (n^{-5})^{-5}(r^{3})^{-5}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(n^{-1})^{-4}}}{{(n^{-5}r^{3})^{-5}}} = \dfrac{{n^{4}}}{{n^{25}r^{-15}}}$ Break up the equation by variable and simplify. $\dfrac{{n^{4}}}{{n^{25}r^{-15}}} = \dfrac{{n^{4}}}{{n^{25}}} \cdot \dfrac{{1}}{{r^{-15}}} = n^{{4} - {25}} \cdot r^{- {(-15)}} = n^{-21}r^{15}$.